This integral requires u-substitution for a rational function with a quadratic denominator raised to a power. Let u = x² + 1, so du = 2x dx, which means 10x dx = 5 du. The integral becomes ∫(5/u²) du = 5∫u^(-2) du = 5(-1/u) = -5/u = -5/(x² + 1). The antiderivative of u^(-2) is -1/u. Choice A has the wrong sign, while choice E represents the derivative rather than the antiderivative. When the numerator is a multiple of the derivative of the base of a power in the denominator, use u-substitution.

This integral uses u-substitution for a rational function with a linear denominator. Let u = 3x + 2, so du = 3 dx, which means 6 dx = 2 du. The integral becomes ∫(2/u) du = 2 ln|u| + C = 2 ln|3x + 2| + C. The coefficient 6 divided by the derivative factor 3 gives 2. Choice A incorrectly multiplies by 6, while choice C omits the adjustment factor. When integrating a constant over a linear expression ax + b, divide the constant by a.

This integral uses u-substitution for a rational function where the numerator is a multiple of the derivative of the denominator. Let u = x² + 1, so du = 2x dx, which means 5(2x) dx = 10x dx becomes ∫(5/u) du = 5 ln|u| + C = 5 ln(x² + 1) + C. Since x² + 1 > 0, we can omit absolute value bars. Choice B uses 5/2 incorrectly, while choice D would result from an arctangent substitution. When the numerator is a multiple of the derivative of the denominator, use logarithmic integration.

This integral requires u-substitution for a rational trigonometric function. The integrand 6sin(3x)/cos(3x) = 6tan(3x) can be rewritten, but it's better to use u = cos(3x), so du = -3sin(3x) dx. This gives us ∫6sin(3x)/cos(3x) dx = ∫(-2/u) du = -2ln|u| = -2ln|cos(3x)|. The factor 6 divided by -3 gives -2. Choice B has the wrong sign, while choice E represents the derivative form. When integrating f'(x)/f(x), use u-substitution to get ln|f(x)| + C.

This integral requires u-substitution to handle the rational function with a quadratic denominator. Let u = t² + 4, so du = 2t dt, which means 6t dt = 3 du. The integral becomes ∫(3/u) du = 3 ln|u| + C. Substituting back gives 3 ln(t² + 4) + C (absolute value bars can be dropped since t² + 4 > 0). Choice D might tempt students who think of arctangent formulas, but that applies to ∫1/(t² + 4) dt, not our integral with t in the numerator. When the numerator contains the derivative of the denominator (up to a constant), use u-substitution with the denominator as u.

This integral uses u-substitution for an exponential function with a negative linear exponent. Let u = -6t, so du = -6 dt, which means dt = -(1/6) du. The integral becomes $∫12e^u$(-1/6) du = $-2∫e^u$ du = $-2e^u$ = -2e^(-6t). The factor 12 combined with -1/6 gives -2. Choice B has the wrong sign, while choice C incorrectly multiplies by -12. For exponential functions with negative exponents, the antiderivative will be negative times the exponential divided by the coefficient.

This integral uses u-substitution for a power function with a quadratic inner expression. Let u = x² + 4x + 5, so du = (2x + 4) dx, making the integral ∫u³ du = u⁴/4 = (1/4)(x² + 4x + 5)⁴. The derivative of the inner function 2x + 4 exactly matches the linear factor in the integrand. Choice B omits the factor of 1/4, while choice C has an incorrect denominator. When the linear factor exactly matches the derivative of a quadratic expression, use u-substitution with the quadratic as u.

This integral uses u-substitution for a function involving a square root with a linear expression inside. Let $u = 1 - 4x$, so $du = -4 , dx$, which means $dx = -\dfrac{1}{4} , du$. The integral becomes $\int 2 u^{-1/2} (-1/4) , du = -\dfrac{1}{2} \int u^{-1/2} , du = -\dfrac{1}{2} (2 \sqrt{u}) = -\sqrt{u} = -\sqrt{1-4x}$. The antiderivative of $u^{-1/2}$ is $2 u^{1/2} = 2 \sqrt{u}$, and we include the negative sign from the substitution. Choice C has an extra factor of 1/2, while choice D would result from an inverse trigonometric substitution. For integrals of the form $\dfrac{f'(x)}{\sqrt{f(x)}}$, use u-substitution with $u = f(x)$.

This integral uses u-substitution for a rational function where the numerator is a multiple of the derivative of the denominator. Let u = x⁴ + 2, so du = 4x³ dx, which means 10x³ dx = (10/4) du = (5/2) du. The integral becomes ∫(5/2)(1/u) du = (5/2) ln|u| + C = (5/2) ln|x⁴ + 2| + C. Choice A incorrectly uses 10 instead of 5/2, while choice C uses 10/4 = 5/2 but writes it as a fraction. When the numerator is a multiple of the derivative of the denominator, divide by the appropriate factor.

This integral uses u-substitution for a rational function with a linear denominator. Let $u = 1 + 2x$, so $du = 2 , dx$, which means $3 , dx = \frac{3}{2} du$. The integral becomes $\int \frac{3}{2} \frac{1}{u} , du = \frac{3}{2} \ln|u| + C = \frac{3}{2} \ln|1 + 2x| + C$. The coefficient 3 divided by the derivative factor 2 gives 3/2. Choice B incorrectly multiplies by 3 instead of dividing by 2, while choice C omits the adjustment factor. When integrating over a linear expression $ax + b$, divide the coefficient by a.